The following code is legal in C++11.
template<int... N>
std::tuple<decltype(N)...> f()
{
return std::make_tuple(7 + N...);
}
What does it mean?
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How do you know it's legal? If you fed it to a compiler and it didn't complain, it means just about nothing. – n. m. could be an AI Jul 09 '14 at 07:49
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Just to clarify: `+` has higher precedence than `...`, so this is the same as `make_tuple( (7 + N) ... )`. – Potatoswatter Jul 09 '14 at 07:51
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1This code is not legal in C++11. This is C++14 code due to the missing trailing return type. – rubenvb Jul 09 '14 at 07:53
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@n.m. gcc uses this technique a lot to implement `std::integer_sequence` – xmllmx Jul 09 '14 at 07:55
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"gcc uses this technique a lot". This tells **nothing** about legality of the code. – n. m. could be an AI Jul 09 '14 at 07:56
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I removed the offending C++14, which doesn't add anything meaningful anyway. – Potatoswatter Jul 09 '14 at 07:59
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3@n.m., I just believe in gcc blindly. – xmllmx Jul 09 '14 at 08:03
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1@Potatoswatter now there is no compilable code in the question, bad edit IMHO... – rubenvb Jul 09 '14 at 08:05
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That's a tad overly optimistic, and worth a mention in the question itself, just in case. – n. m. could be an AI Jul 09 '14 at 08:07
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1There, made it C++11. – rubenvb Jul 09 '14 at 08:08
1 Answers
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First of all, look at the template parameters: template <int ... N>. Even though a variable number of template arguments can be given to f, all of them must be of type int.
Now when you use f<t1, t2, ..., tn>, the parameter unpacking (7 + N...) will follow the pattern 7 + N and expand to
7 + t1, 7 + t2, 7 + t3, ..., 7 + tn
Therefore you end up with a tuple which contains each of your template arguments increased by seven. The details can be found in section 14.5.3 Variadic templates [temp.variadic].
3. A pack expansion consists of a pattern and an ellipsis, the instantiation of which produces zero or more instantiations of the pattern in a list [...].
Zeta
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1[Live example](http://coliru.stacked-crooked.com/a/239fe332227e245b). Note the `-std=c++1y`. – rubenvb Jul 09 '14 at 07:52
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@rubenvb: Noticed it beforehand, but didn't think about the automatic type deduction that much (too much Haskell lately). But now that I think about it, is it actually possible to create a function with the same template parameters and type in C++11 without sacrificing kittens? – Zeta Jul 09 '14 at 08:11
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@rubenvb: I see myself out. I knew there was something like `decltype`, but I haven't used templates meta programming in ages. Two hours of sleep isn't enough anymore :D. – Zeta Jul 09 '14 at 08:16